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Simple Stress and Strain 
Lecturer; 
Dr. Dawood S. Atrushi
Content 
¢ Stress 
¢ Strain (Deformation) 
¢ Stress-Strain relationship 
¢ Hook’s law 
¢ Poisson’s ration 
¢ Thermal...
STRESS 
Chapter Chapter 2 2 Stress Stress and and Strain 
Strain 
External and internal forces in a structural member 
2.0...
Definition of Stress 
In the previous figures; 
l ΔA infinitesimal regions 
l ΔF distributed/resultant force on ΔA 
l Δ...
look at the whole sectioned area, we can say number of forces, we Stress can define expression 
where each one (of magnitu...
and carried by the material and the deformation they create. 
STRESS (SI&4th p. 22-23, 3rd p. 22-23) 
Normal and Shear Str...
we need to consider both magnitude and direction Now let’s resolve the force 'F in normal and tangential 2.1b). The intens...
Now let’s resolve the force 'F in normal and 2.1b). The intensity of the force or force per unit Normal Stress, V (sigma),...
“pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expres...
mechanical analysis of deformable solids 
force system 
distributed along a 
continuous surface ≡ Mechanics of materials o...
• The loaded perpendicular • The the • The resultant of the internal forces for an axially 
loaded member is normal to a s...
Measurement of mechanical (strength) properties 
Uniaxial Compressive 
Test 
Measurement of mechanical (strength) properti...
Units of Stress 
¢ The units in the SI system is the 
Newton per square meter or Pascal, 
i.e. : Pa = N/m2. 
¢ In engine...
Example 1 
An 80 kg lamp is 
supported by a 
single electrical 
copper cable of 
diameter d = 3.15 
mm. What is the 
stres...
Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=Giga Pascal GPa (=Pau109) e.g. 20,000,...
Review of Static 
Example 2 
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WRVXSSRUWDN1 ORDG 
• 7KHVWUXFWXUHFRQVLVWVRID 
ERRPDQGURGMRLQHGE 
SL...
DWWKH 
MXQFWLRQVDQGVXSSRUWV 
• 3HUIRUPDVWDWLFDQDOVLV 
WRGHWHUPLQHWKHLQWHUQDO 
IRUFHLQHDFKVWUXFWXUDO 
PHPEHUDQGWKHUHDFWLRQ ...
Deformation 
¢ Whenever a force is applied to a body, 
its shape and size will change. These 
changes are referred as def...
¢ It is however very hard to make a 
relative comparison between bodies 
or structures of different size and 
length as t...
F F 
(+) () 
Original length 
Deformation 
(elongation) 
F F 
Deformation 
(contraction) 
Deformation 
(contraction) 
Orig...
(a) Before deformed Fig. 2.3 Generalized deformation due Let’s take the arbitrarily shaped body in Fig. 2.3 as an line seg...
and B are displaced to A’ and B’ and in general the The change in length of the line is therefore 'S-'S’. strain Average m...
Unit of Strain 
¢ The normal strain is a dimensionless quantity 
since it is a ratio of two lengths, but it is common in ...
Example 3 
¢ If it is measured that the a cable was 
elongated by 1.35 mm due to the 
weight of the light, what would its...
Stress-Strain Relationship 
Material Test and Stress-Strain Diagram 
¢ The material strength depends on its 
ability to s...
2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI4th p.83-85-95) 
Material Test and Stress-Strain Diagram 
The material stre...
Hooke’s Law 
¢ The stress-strain linear relationship was 
discovered by Robert Hook in 1676 and is 
known as Hooke's law....
When the load P is applied to the bar it changes strain in axial direction and in lateral/radial direction axial Poisson’s...
direction of force and yet its sides expand laterally as in Fig. 2.5(b). 
When the load P is applied to the bar it changes...
terms as the Modulus of Elasticity or Young's Modulus with units engineering metal material, GPa is used, e.g. mild steel ...
Likewise, a compressive force acting on a deformable direction of force and yet its sides expand laterally as in When the ...
Remarks 
¢ The lateral strain is caused only by 
axial force. No force or stress acts in 
lateral direction; 
¢ Lateral ...
Strain in Lateral Direction 
¢ For bars subjected to a tensile stress 
σx, the strains in the y and z planes 
are: 
­ 
H ...
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Simple Stress and Strain

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Simple Stress and Strain

  1. 1. Simple Stress and Strain Lecturer; Dr. Dawood S. Atrushi
  2. 2. Content ¢ Stress ¢ Strain (Deformation) ¢ Stress-Strain relationship ¢ Hook’s law ¢ Poisson’s ration ¢ Thermal or Temperature stress and strain Wednesday, 2 Strenght of Materials I - DAT October 22, 14
  3. 3. STRESS Chapter Chapter 2 2 Stress Stress and and Strain Strain External and internal forces in a structural member 2.0 INTRODUCTION In the first Chapter we discuss the equations of statics, and how to determine the ground reaction for any structure. The method can also be used to determine the internal loads carried by the members or parts of a body. We now need to define how these internal loads are distributed and carried by the material and the deformation they create. 2.1 STRESS (SI&4th p. 22-23, 3rd p. 22-23) Consider an element of continuous (no voids) and cohesive (no cracks, breaks and defects) material subjected to a number of externally applied loads as shown in Fig. 2.1a). It is supposed that the member is in equilibrium. 2.0 INTRODUCTION In the first Chapter we discuss the equations of statics, and how to determine the ground reaction for any structure. The method can also be used to determine the internal loads carried by the members or parts of a body. We now need to define how these internal loads are distributed ¢ Consider and carried by the material an element and the deformation of they continuous create. 2.1 STRESS and (SI&cohesive 4th p. 22-23, 3rd p. 22-material 23) subjected to a Consider number an element of continuous of externally (no voids) and cohesive (no cracks, breaks material subjected to a number of externally applied applied loads as shown loads. and defects) in Fig. 2.1a). It is supposed that the member is in equilibrium. Free Body Diagram Free Body Diagram 'F n 'Fn 'F 'Ft F4 F4 F3 F5 'Fn F3 F5 Cross section: A 'A Cross section: A 'A F1 F2 t t F1 F2 n F1 F2 (a) (b) (a) (b) Fig. 2.1 External and internal forces in a structural member 'Ft F2 F1 Fig. 2.1 External and internal forces in a structural member Wednesday, 3 Strenght of Materials I - DAT October 22, 14 If we now cut this body, the applied forces can be thought of as being distributed over the cut area A as in Fig. 2.1b). Now if we look at infinitesimal regions 'A, we assume the resultant If we now cut this body, the applied forces can be thought of as being distributed over the cut area A as in Fig. 2.1b). Now if we look at infinitesimal regions 'A, we assume the resultant
  4. 4. Definition of Stress In the previous figures; l ΔA infinitesimal regions l ΔF distributed/resultant force on ΔA l ΔA is extremely small, ΔF is uniform l Entire area A is subject to an infinite number of forces, each one acts over a small area ΔA. Stress is the intensity of the internal force on a specific plane passing through a point. Wednesday, 4 Strenght of Materials I - DAT October 22, 14
  5. 5. look at the whole sectioned area, we can say number of forces, we Stress can define expression where each one (of magnitude stress. Definition: Stress is the intensity of the ¢ Mathematically, passing stress through can be a point. Mathematically, expressed as; stress can be expressed as lim F A ' A ' V ' o0 Dividing the magnitude of internal force 'F by let ¢ If we let ΔA approach zero, we obtain the stress at a point. 'A approach zero, we obtain the stress at body, which depends on the position that important concepts that we introduced in mechanics 5 Strenght of Materials I - DAT Wednesday, October 22, 14 Normal and Shear Stress
  6. 6. and carried by the material and the deformation they create. STRESS (SI&4th p. 22-23, 3rd p. 22-23) Normal and Shear Stress an element of continuous (no voids) and cohesive (no cracks, breaks and defects) subjected to a number of externally applied loads as shown in Fig. 2.1a). It is that the member is in equilibrium. Normal Stress, σ ¢ Now let’s resolve the force ΔF in normal and tangential direction of the acting area. The intensity of the force or force per unit area acting normally to section A is called Normal Stress 'F 'Fn 'Ft F4 Cross section: A 'A F1 Wednesday, 6 Strenght of Materials I - DAT October 22, 14 F2 F3 F5 Free Body Diagram F1 F2 t n (a) (b) Fig. 2.1 External and internal forces in a structural member cut this body, the applied forces can be thought of as being distributed over the cut
  7. 7. we need to consider both magnitude and direction Now let’s resolve the force 'F in normal and tangential 2.1b). The intensity of the force or force per unit Normal Stress, V (sigma), and it is expressed as: ¢ Normal stress, σ (sigma) is expressed as: F ' lim n A ' V ' o0 A If this stress “pulls” on the area it is referred as “pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as: ¢ If this stress “pulls” on the area it is referred as Tensile Stress and defined as Positive. ¢ If it “pushes” on the area it is called Compressive Stress Negative. ' F and defined as lim t A ' W ' o0 A 7 Strenght of Materials I - DAT Wednesday, October 22, 14 Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-
  8. 8. Now let’s resolve the force 'F in normal and 2.1b). The intensity of the force or force per unit Normal Stress, V (sigma), and it is expressed as: F ' lim n A ' V A If this stress “pulls” on the area it is referred “pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as: Shear Stress, τ ' o0 ¢ The intensity or force per unit area acting tangentially to A is called Shear Stress, τ (tau), and it is expressed as: F ' lim t A ' W ' o0 A Average Normal Stress (SI&4th p. 24-31, 3rd p. To begin, we only look at beams that carry tensile and slender. Such beams can then be assumed Wednesday, 8 Strenght of Materials I - DAT October 22, 14
  9. 9. “pushes” on the area it is called Compressive Stress The intensity or force per unit area acting tangentially it is expressed as: Average Normal Stress F ' lim t A ' W Average Normal Stress (SI&4th p. 24-31, 3rd p. 27-To begin, we only look at beams that carry tensile and slender. Such beams can then be assumed to simplified to: ¢ Let us look at beams ' o0 that A carry tensile or compressive loads and which are long and slender. Such beams can then be assumed to carry a constant stress, and the Equation can be simplified to: V F A We call this either Average Normal Stress or Uniform Units of Stress The units in the SI system is the Newton per square In engineering, Pa seems too small, so we usually Kilo Pascal KPa (=Pau103) ¢ We call this either Average Normal Stress or Uniform Uniaxial Stress. Wednesday, 9 Strenght of Materials I - DAT October 22, 14
  10. 10. mechanical analysis of deformable solids force system distributed along a continuous surface ≡ Mechanics of materials of structures: Mechanics of materials of structures: 1. Stress analysis → whether or not internal forces 1. Stress analysis → whether or not internal forces (stresses) due to external STATICS: STRENGTH OF MATERIALS: (stresses) due to external loading cause a) Statics b) Strength of materials STATICS: STRENGTH OF MATERIALS: Force system distributed along a continuous surface force system distributed along a continuous surface Wednesday, 10 Strenght of Materials I - DAT October 22, 14
  11. 11. • The loaded perpendicular • The the • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress. • A uniform distribution of stress in a section infers that the line of action for the resultant the internal forces passes through the centroid of the section. • • A uniform distribution of stress is only possible if the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as centric loading. • If a two-force member is eccentrically loaded, then the resultant of the The Civil and Planning Faculty stress of Engineering distribution Engineering Education – State University in of • The normal stress at a particular ' F point may not be equal to the average stress but the resultant of the stress distribution must satisfy V V P V aveA dF V dA P Wednesday, 11 Strenght of Materials I - DAT October 22, 14 The equal stress • The indeterminate, alone. Internal forces for an axially loaded member • The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy ³ ³ A A A ave A ' ' o 0 lim • The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone.
  12. 12. Measurement of mechanical (strength) properties Uniaxial Compressive Test Measurement of mechanical (strength) properties Uniaxial compressive test of a concrete block: Uniaxial compressive test of a concrete block: http://www.muszakiak.com/diagnosztika-es-anyagvizsgalat/beton-szilardsagi-vizsgalatok.html http://metrumkft.hu/termekek/betonszilardsagvizsgalo F F Wednesday, 12 Strenght of Materials I - DAT October 22, 14 http://www.muszakiak.com/diagnosztika-es-anyagvizsgalat/beton-szilardsagi-vizsgalatok.html F F
  13. 13. Units of Stress ¢ The units in the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2. ¢ In engineering, Pa seems too small, so we usually use: Wednesday, 13 Strenght of Materials I - DAT October 22, 14 Stress (ı) material is subjected to an external force, a resisting force is set up within the component. internal resistance force per unit area acting on a material or intensity of the forces distributed given section is called the stress at a point. It uses original cross section area of the specimen and also known as engineering stress conventional stress. Therefore, P A T P is expressed in Newton (N) and A, original area, in square meters (m2), the stress ǔ will expresses in N/ m2. This unit is called Pascal (Pa). As Pascal is a small quantity, in practice, multiples of this unit is used. 1 kPa = 103 Pa = 103 N/ m2 (kPa = Kilo Pascal) 1 MPa = 106 Pa = 106 N/ m2 = 1 N/mm2 (MPa = Mega Pascal) 1 GPa = 109 Pa = 109 N/ m2 (GPa = Giga Pascal) take an example: A rod 10 mm q10 mm cross-section is carrying an axial tensile load this rod the tensile stress developed is given by 3 10 10 10 P kN q N
  14. 14. Example 1 An 80 kg lamp is supported by a single electrical copper cable of diameter d = 3.15 mm. What is the stress carried by the cable. A d 0.00315 u m F 784 100 6 V a a 80kg F F Section a-a A d FBD Wednesday, 14 Strenght of Materials I - DAT October 22, 14 Stress the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2. engineering, Pa seems too small, so we usually use: Pascal KPa (=Pau103) e.g. 20,000Pa=20kPa Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=20MPa Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=20GPa 2.1: An 80 kg lamp is supported by a single electrical of diameter d = 3.15 mm. What is the stress carried determine the stress in the wire/cable as Eq. (2.4), we need sectional area A of the cable and the applied internal 6 2 2 2 7.793 10 4 S S 4 u F mg 80u9.8 784N . MPa A . 7 793 10 6 u Stress (SI4th p. 48-49, 3rd p. 51-52) Example 2.1, we may concern whether or not 80kg would be too heavy, or say stress would be too high for the wire/cable, from the safety point of view. Indeed, of most important indicators of structural strength. When the stress (intensity of
  15. 15. Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical copper cable of diameter d = 3.15 mm. What is the stress carried by the cable. To determine the stress in the wire/cable as Eq. (2.4), we need the area A of the cable and the applied internal force Pascal MPa (=Pau106) e.g. 20,000,000Pa=Giga Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical copper cable of diameter d = 3.15 mm. What is the stress carried by the cable. To determine the stress in the wire/cable as Eq. (2.4), we need the cross sectional area A of the cable and the applied internal force F: Allowable Stress (SI4th p. 48-49, 3rd p. 51-52) From Example 2.1, we may concern whether or not 80kg 100.6MPa stress would be too high for the wire/cable, from the stress is one of most important indicators of structural strength. Mega In engineering, Pa seems too small, so we usually use: Allowable Stress (SI4th p. 48-49, 3rd p. 51-52) From Example 2.1, we may concern whether or not 80kg 100.6MPa stress would be too high for the wire/cable, from Kilo Pascal KPa (=Pau103) e.g. 20,000Pa=Solution Mega Pascal MPa (=Pau106) e.g. 20,000,000Pa=Example Giga 1 Pascal GPa (=Pau109) e.g. 20,000,000,000Pa=Example 2.1: An 80 kg lamp is supported by a single electrical copper cable The cross of diameter sectional d = 3.15 area mm. A What of the is the stress carried ¢ by the cable. To determine cable the and stress the in applied the wire/internal cable as Eq. (2.4), we need the cross force sectional F: area A of the cable and the applied internal force F: A d u m 6 2 2 2 7.793 10 0.00315 2 2 0.00315 2 2 A d 0.00315 4 u m S S S S 4 u 4 4 S u u F mg 80u9.8 784N 7.793 u 10 m 7.793 10 4 4 F mg 80u9.8 784N F mg 80u9.8 784N F 784 100 6 F 784 100 6 so . MPa V 6 2 so . MPa V F A 7 . 793 784 10 100 6 6 u 6 2 so . MPa V A . 6 7 793 10 6 u A . 7 793 u 10 Allowable Stress (SI4th p. 48-49, 3rd p. 51-52) From 15 Example 2.1, Strenght we may of Materials concern I - DAT whether Wednesday, or October not 22, 80kg 14 would 100.6MPa would be too high for the wire/cable, from the safety stress is one of most important indicators of structural strength. When
  16. 16. Review of Static Example 2 • 7KHVWUXFWXUHLVGHVLJQHG WRVXSSRUWDN1 ORDG • 7KHVWUXFWXUHFRQVLVWVRID ERRPDQGURGMRLQHGE SLQV]HURPRPHQW FRQQHFWLRQV
  17. 17. DWWKH MXQFWLRQVDQGVXSSRUWV • 3HUIRUPDVWDWLFDQDOVLV WRGHWHUPLQHWKHLQWHUQDO IRUFHLQHDFKVWUXFWXUDO PHPEHUDQGWKHUHDFWLRQ IRUFHVDWWKHVXSSRUWV Wednesday, 16 Strenght of Materials I - DAT October 22, 14 The Civil and Planning Engineering Education Department Faculty of Engineering – State University of Yogyakarta
  18. 18. Deformation ¢ Whenever a force is applied to a body, its shape and size will change. These changes are referred as deformations. ¢ Positive (elongation) or negative (contraction) deformation; 2.2 DEFORMATION (SI4th p. 67-68, 3rd p. 70) Whenever a force is applied to a body, its shape and size will change. These changes referred as deformations. These deformations can be thought of being either positive (elongation) or negative (contraction) in sign as shown in Fig. 2.2. DEFORMATION (SI4th p. 67-68, 3rd p. 70) Whenever a force is applied to a body, its shape and size will change. These changes are referred as deformations. These deformations can be thought of being either positive elongation) or negative (contraction) in sign as shown in Fig. 2.2. F F F F F (+) () (+) () Original length Deformation (elongation) F Deformation Deformation (contraction) (contraction) Original length Deformation (elongation) Original length Fig. 2.2 Deformation due to applied axial forces Original length Fig. 2.2 Deformation due to applied axial forces Wednesday, 17 Strenght of Materials I - DAT October 22, 14 It is however very hard to make a relative comparison between bodies or structures of different however very hard to make a relative comparison between bodies or structures of different
  19. 19. ¢ It is however very hard to make a relative comparison between bodies or structures of different size and length as their individual deformations will be different. This requires the development of the concept of Strain, which relates the body’s deformation to its initial length. Wednesday, 18 Strenght of Materials I - DAT October 22, 14
  20. 20. F F (+) () Original length Deformation (elongation) F F Deformation (contraction) Deformation (contraction) Original length Original length Strain Deformation (elongation) Normal Strain ¢ The elongation (+ve) or contraction (−ve) of a body per unit length is termed Strain. Original length Fig. 2.2 Deformation due to applied axial forces Fig. 2.2 Deformation due to applied axial forces It is however very hard to make a relative comparison between bodies or structures of different size and length as their individual deformations will be different. This requires the development of the concept of Strain, which relates the body’s deformation to its initial length. 2.3 STRAIN (SI p.68-69; 4th p. 67-69; 3rd p.71-72) Normal Strain The elongation (+ve) or contraction (ve) of a body per unit length is termed Strain. It is however very hard to make a relative comparison between bodies or structures of different size and length as their individual deformations will be different. This requires the development of the concept of Strain, which relates the body’s deformation to its initial length. 2.3 STRAIN (SI p.68-69; 4th p. 67-69; 3rd p.71-72) Normal Strain The elongation (+ve) or contraction (ve) of a body per unit length is termed Strain. n A 'S B A’ A 'S B A’ B’ F1 'S’ B’ 'S’ F2 (a) Before deformed (b) After deformed F3 (a) Before deformed (b) After deformed Fig. 2.3 Generalized deformation due to applied forces Wednesday, 19 Strenght of Materials I - DAT October 22, 14 F3 F2 F1 n Fig. 2.3 Generalized deformation due to applied forces Let’s take the arbitrarily shaped body in Fig. 2.3 as an example. Consider the infinitesimal line segment AB that is contained within the undeformed body as shown in Fig. 2.3(a). The Let’s take the arbitrarily shaped body in Fig. 2.3 as an example. Consider the infinitesimal
  21. 21. (a) Before deformed Fig. 2.3 Generalized deformation due Let’s take the arbitrarily shaped body in Fig. 2.3 as an line segment AB that is contained within the undeformed line AB lies along the n-axis and has an original length ¢ After deformation, the generalized and B are strain displaced mathematically to A’ and B’ can and be in general the line The change expressed in length as; of the line is therefore 'S-'S’. We strain mathematically as ' ' lim S' S B A ' H (BoA along n) o S Average Normal Strain If the stress in the body is everywhere constant, in other the material (e.g. uniform uniaxial tension or compression) be computed 20 by Strenght of Materials I - DAT Wednesday, October 22, 14 L L
  22. 22. and B are displaced to A’ and B’ and in general the The change in length of the line is therefore 'S-'S’. strain Average mathematically Normal as Strain ' ' lim S' S B A ' If the H stress in the body (Bis oeverywhere A along n) ¢ o S constant, in other words, the Average deformation Normal Strain is uniform in the material If the stress (e.g. in uniform the body uniaxial is everywhere tension or constant, in other the material compression), (e.g. uniform the uniaxial strain ε tension can be or compression) be computed computed by by; L Deformed Original ' L i.e. the change in length of the body over its original L L H L Original Wednesday, 21 Strenght of Materials I - DAT October 22, 14
  23. 23. Unit of Strain ¢ The normal strain is a dimensionless quantity since it is a ratio of two lengths, but it is common in practice to state it in terms of a ratio of length units. i.e. meters per meter (m/m) ¢ Usually, for most engineering applications ε is very small, so measurements of strain are in micrometers per meter (μm/m) or (μ/m). ¢ Sometimes for experiment work, strain is expressed as a percent, e.g. 0.001m/m = 0.1%. A normal strain of 480μm for a one-meter length is said: ε= 480×10-6 = 480(μm/m) = 0.0480% = 480μ (micros) = 480μs (micro strain) Wednesday, 22 Strenght of Materials I - DAT October 22, 14
  24. 24. Example 3 ¢ If it is measured that the a cable was elongated by 1.35 mm due to the weight of the light, what would its strain be? Wednesday, 23 Strenght of Materials I - DAT October 22, 14
  25. 25. Stress-Strain Relationship Material Test and Stress-Strain Diagram ¢ The material strength depends on its ability to sustain a load without undue deformation or failure. ¢ The property is inherent in the material itself and must be determined by experiment. ¢ The tests are performed in universal test machine. Wednesday, 24 Strenght of Materials I - DAT October 22, 14
  26. 26. 2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI4th p.83-85-95) Material Test and Stress-Strain Diagram The material strength depends on its ability to sustain a load without undue deformation failure. The property is inherent in the material itself and must be determined by experiment. One of the most important tests to perform in this regard is the tension or compression. so, a bunch of standard specimen is made. The test is performed in universal test machine. Shown in Fig. 2.4 is the specimen and test result of Stress-Strain Diagram. 2.4 STRESS-STRAIN RELATIONSHIP, HOOKE'S LAW (SI4th p.83-93; p.85-95) Material Test Material and Stress-Strain The material strength depends test Diagram on its ability to sustain a load without undue deformation failure. The property is inherent in the material itself and must be determined by experiment. One of the most Stress-important Strain tests to perform Diagram in this regard is the tension or compression. To so, a bunch of standard specimen is made. The test is performed test machine. Shown in Fig. 2.4 is the specimen and test result of Stress-Strain F Standard Specimen Yield stress F V V Elastic Yielding Hardening VY VY Ultimate stress Vu Fracture stress Vf Plastic behavior Elastic behavior ProportionalVpl limit Fig. 2.4 Material test and Stress-Strain Diagram Wednesday, 25 Strenght of Materials I - DAT October 22, 14 H Elastic Yielding Hardening Necking Standard Specimen Yield stress Ultimate stress Vu Fracture stress Vf Plastic behavior Elastic behavior ProportionalVpl limit Fig. 2.4 Material test and Stress-Strain Diagram F F H The Stress-Strain diagram consists of 4 stages during the whole process, elastic, yielding, hardening and necking stages respectively. From yielding stage, some permanent plastic deformation occurs. About 90% of engineering problems only concern the elastic deformation The Stress-Strain diagram consists of 4 stages during the whole process, elastic, yielding, hardening and necking stages respectively. From yielding stage, some permanent
  27. 27. Hooke’s Law ¢ The stress-strain linear relationship was discovered by Robert Hook in 1676 and is known as Hooke's law. ¢ It is mathematically represented by; V EH where E is terms as the Modulus of Elasticity For l E most is Modulus of engineering of Elasticity or metal Young's Modulus with units of N/m2 or Pa (material, or GPa GPa) . For mild steel it is about 2.5 200GPa~POISSON'S 210GPa. RATIO (SI4th Ed Definition Wednesday, 26 Strenght of Materials I - DAT October 22, 14
  28. 28. When the load P is applied to the bar it changes strain in axial direction and in lateral/radial direction axial Poisson’s Ratio ¢ When a deformable body is stretched by a tensile force, not only does it elongate but it also contract laterally, i.e. it would contract in other two dimensions. v Lateral Strain H Wednesday, 27 Strenght of Materials I - DAT October 22, 14 G L H In early 1800s, French scientist Poisson realized two strains is a constant. We called the constant Lateral Axial Strain H Axial Original shape Defrormed shape F F r Gr L L+G (a) Fig. 2.5 Relationship of the axial The negative sign is used here since longitudinal contraction (negative strain), vice versa. So Poisson’s
  29. 29. direction of force and yet its sides expand laterally as in Fig. 2.5(b). When the load P is applied to the bar it changes the bar’s length by G and its radius by Gr G H ¢ Likewise, a compressive force acting on a deformable body cause it to contract in the direction of force and yet its sides expand laterally r H Lateral v Lateral Strain (2.11) L Gr Wednesday, 28 Strenght of Materials I - DAT October 22, 14 . The strain in axial direction and in lateral/radial direction are respectively axial L r lateral G H In early 1800s, French scientist Poisson realized that within elastic range the ratio of these two strains is a constant. We called the constant as Poisson’s ratio by v (Nu) Axial Axial Strain H Original shape Defrormed shape F F r Gr L L+G Original shape Defrormed shape F F L – G r (a) (b) Fig. 2.5 Relationship of the axial strain with the lateral strain The negative sign is used here since longitudinal elongation (positive strain) cause lateral contraction (negative strain), vice versa. So Poisson’s ration is positive, i.e. vt0.
  30. 30. terms as the Modulus of Elasticity or Young's Modulus with units engineering metal material, GPa is used, e.g. mild steel is about 200GPa~POISSON'S RATIO (SI4th Ed p104-105; 3rd Ed p107-108) deformable body ¢ is When stretched the by load a tensile P is applied force, not to only the bar does it elongate laterally, i.e. it would it changes contract the in bar’s other length two dimensions by δ and as shown compressive force its radius acting by on δr. a The deformable strain in body axial cause it to force and yet its sides expand laterally as in Fig. 2.5(b). load P is applied to direction the and in lateral/radial direction direction and are in lateral/respectively; bar it changes the bar’s length by G and its radius radial direction are respectively G H axial L G r r H lateral 1800s, French scientist Poisson realized that within elastic range the a constant. We called the constant as Poisson’s ratio by v (Nu) Wednesday, 29 Strenght of Materials I - DAT October 22, 14 H Lateral Lateral Strain H
  31. 31. Likewise, a compressive force acting on a deformable direction of force and yet its sides expand laterally as in When the load P is applied to the bar it changes the bar’s strain in axial direction and in lateral/radial direction are G H axial ¢ Within the elastic range, the ratio of these two strains is a constant, L and it is called Poisson’s ratio by v (Nu); v Lateral Strain H ¢ Poisson’s ration is positive, i.e. v≥0. Wednesday, 30 Strenght of Materials I - DAT October 22, 14 H lateral In early 1800s, French scientist Poisson realized that two strains is a constant. We called the constant as Poisson’s Lateral Axial Strain H Axial Original shape Defrormed shape F F
  32. 32. Remarks ¢ The lateral strain is caused only by axial force. No force or stress acts in lateral direction; ¢ Lateral strain is the same in all lateral direction; ¢ Usually 0 ≤ v ≤ 0.5 . For most linearly elastic material v=0.3; ¢ Poisson’s ratio is a constant. Wednesday, 31 Strenght of Materials I - DAT October 22, 14
  33. 33. Strain in Lateral Direction ¢ For bars subjected to a tensile stress σx, the strains in the y and z planes are: ­ H °°° V H H ® °°° H H ¯ x E v v x y x V v v E V E x x z x 2.6 THERMAL STRAIN (SI4th Ed p. 148-152; 3rd 32 Strenght of Materials I - DAT Wednesday, October 22, 14 Thermal Deformation When the temperature of a body is changed, its overall
  34. 34. Wednesday, 33 Strenght of Materials I - DAT October 22, 14

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